On permutations of left cosets Let $K$ be a subgroup of some group $H$; let $X$ be the set of left cosets of $K$, i.e. $X = \{hK: h \in H\}$; and let $G$ be the group of permutations of $X$.  For all $h \in H$, let $f\,(h) \in G$ be the permutation of $X$ that sends every coset $h'K$ to the coset $hh'K$.  It's easy to see that the map $f:H\to G$ is a homomorphism of groups.  My question is

what is the kernel of $f\;\;$?

(I understand that $K\subseteq \ker(f\,)$, and also that if $K$ is contained in the center of $H$, then $\ker(f\,) = K$, but I'm interested in the general case.)
Thanks!
 A: Just to add to m.k.'s response, the $Ker(f)=\bigcap_{a \in H} a^{-1}Ka=\text{core}_H(K)$, called the normal core of $K$ in $H$, and is the largest normal subgroup of H that is contained in K, see this for more information.
I am curious though kjo why you are asking this particular question, because I did some work related to your question.  If the group $H$ is the symmetry group of some set of objects acting transitively, one can talk about a coloring of the objects by assigning a unique color to each left coset of $K$.  If I am not mistaken, $G$ is in fact, just $H$, called in this setting the color symmetry group of the coloring, and the $ker(f)$ is called the colored fixing group.  
A: It is not true in general that $K$ is contained in the kernel of $f$. The kernel contains $K$ if and only if it equals $K$, because
$Ker(f) = \{h \in H: haK = aK \text{ for every } a \in H \} = \bigcap_{a \in H} a^{-1}Ka$
and so $Ker(f) \leq K$.
For example, let $H = S_3$ and $K = \{(1), (12)\}$. Then $K$ has cosets $C_1 = \{(1), (12)\}$, $C_2 = \{(13),(123)\}$ and $C_3 = \{(23), (132)\}$. Now the permutation $f((12))$ maps $C_2$ to $C_3$, so it is not the identity. Thus $(12) \notin Ker(f)$. In particular, $K$ is not contained in $Ker(f)$.
So although $kK = K$ for every $k \in K$, it is not necessarily true that $kaK = aK$ for arbitrary $a \in H$. But it is true if (and only if) $K$ is normal in $H$.
A: 
After seeing m.k.'s proof, I realized that
$$h \in \ker(f\,) \Leftrightarrow h1_HK = 1_HK \Leftrightarrow hK = K \Leftrightarrow h \in K.$$

